Add finished ch03 exercises.

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Kyle Isom 2018-01-17 11:16:30 -08:00
parent 6d4fb673ef
commit 5dd0bfcf19
2 changed files with 53 additions and 1 deletions

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@ -140,6 +140,14 @@ yes
no no
``` ```
Answer (in `succ.pl`):
```
greater_than(succ(succ(X)), succ(0)).
greater_than(succ(X), succ(Y)) :-
greater_than(X, Y).
```
#### Exercise 3.5 #### Exercise 3.5
Binary trees are trees where all internal nodes have exactly two children. The Binary trees are trees where all internal nodes have exactly two children. The
@ -159,4 +167,6 @@ tree that is its first argument. For example:
?- swap(tree(tree(leaf(1), leaf(2)), leaf(4)),T). ?- swap(tree(tree(leaf(1), leaf(2)), leaf(4)),T).
T = tree(leaf(4), tree(leaf(2), leaf(1))). T = tree(leaf(4), tree(leaf(2), leaf(1))).
yes yes
``` ```
This one took a lot of thinking, but the solution is in `tree.pl`.

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lpn/ch03/tree.pl Normal file
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%% Exercise 3.5
%%
%% Binary trees are trees where all internal nodes have exactly two
%% children. The smallest binary trees consist of only one leaf node. We
%% will represent leaf nodes as `leaf(Label)`. For instance, `leaf(3)`
%% and `leaf(7)` are leaf nodes, and therefore small binary trees. Given
%% two binary trees B1 and B2 we can combine them into one binary tree
%% using the functor `tree/2` as follows: `tree(B1,B2)`. So, from the
%% leaves `leaf(1)` and `leaf(2)` we can build the binary tree
%% `tree(leaf(1),leaf(2))` . And from the binary trees
%% `tree(leaf(1),leaf(2))` and `leaf(4)` we can build the binary tree
%% `tree(tree(leaf(1), leaf(2)),leaf(4))`.
%%
%% Now, define a predicate `swap/2`, which produces the mirror image of
%% the binary tree that is its first argument. For example:
%%
%% ```
%% ?- swap(tree(tree(leaf(1), leaf(2)), leaf(4)),T).
%% T = tree(leaf(4), tree(leaf(2), leaf(1))).
%% yes
%% ```
%% Let's start with the base case: swapping a leaf is the identity.
swap(leaf(X), leaf(X)).
%% What does it mean to swap a tree? Given a tree
%% tree(X, Y)
%% it should have an equivalent
%% tree(swap(Y), swap(X))
%% How do we express this?
%% Let's start with a tree made of two leaves.
%% swap(tree(X, leaf(Y)), tree(leaf(Y), Z).
%% But this isn't quite right, is it? We need to express that
%% the X node needs to be swapped. So we need to express this
%% as a rule.
swap(tree(X, leaf(Y)), tree(leaf(Y), Z)) :- swap(X, Z).
%% And there it is:
%% ?- swap(tree(tree(leaf(1), leaf(2)), leaf(4)),T).
%% T = tree(leaf(4), tree(leaf(2), leaf(1))).